Q. 37 L5.0( 1 Vote )

# Solve each of the

, y = 2 when

Given and

This is a first order linear differential equation of the form

Here, P = –3 cot x and Q = sin 2x

The integrating factor (I.F) of this differential equation is,

We have

[ m log a = log am]

I.F = cosec3x [ elog x = x]

Hence, the solution of the differential equation is,

Recall

ycosec3x = 2(–cosec x) + c

ycosec3x = –2cosec x + c

y = –2sin2x + csin3x

However, when, we have y = 2

2 = –2(1)2 + c(1)3

2 = –2 + c

c = 4

By substituting the value of c in the equation for y, we get

y = –2sin2x + (4)sin3x

y = –2sin2x + 4sin3x

Thus, the solution of the given initial value problem is y = –2sin2x + 4sin3x

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